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Feliz [49]
3 years ago
14

A student buys a rope at the store. The label on the packaging says that the rope is 2.15 meters in length. The student measures

the rope as 1.85m. What is the student’s percent error?
Chemistry
1 answer:
Naily [24]3 years ago
3 0

Answer:

30 center meters

Explanation:

2.15-1.85=0.30

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For each metal complex give the coordination number for the metal species. m(nh34cl2
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The coordination number is 6. The outer Cl- ion isn't included, so you count 4 x NH3 and 2 x Cl- are within Co's coordination sphere.
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Sometimes, during the DNA replication process, mistakes are made. These are called
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What is the hybridization of the central atom, electron geometry, and molecular geometry of H3CCCH?
DIA [1.3K]

Answer:

Hybridization: sp

Electron geometry: linear

Molecular geometry: linear

Explanation:

H₃CCCH can also be written as its Lewis structure which is shown in the figure attached. The figure shows that the central carbon atom makes a single bond with CH₃ and a triple bond with CH. This means that the hybridization of the carbon is sp and both the electron and molecular geometry are linear with an 180° bond angle.

6 0
3 years ago
Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
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